If ABCDEF is a regular hexagon then `vec(AD)+vec(EB)+vec(FC)` equals :

Options

A. (a)2AB
B. (b)3AB
C. (c)4AB
D. (d)none of these

Explanation

To solve the problem, we need to find the sum of the vectors \( \vec{AD} + \vec{EB} + \vec{FC} \) for a regular hexagon ABCDEF.

### Step-by-Step Solution:

1. **Understanding the Regular Hexagon**:
A regular hexagon has equal sides and angles. The vertices can be denoted as A, B, C, D, E, and F.

2. **Identifying the Vectors**:
- \( \vec{AD} \) is the diagonal from A to D.
- \( \vec{EB} \) is the diagonal from E to B.
- \( \vec{FC} \) is the diagonal from F to C.

3. **Using Properties of the Hexagon**:
In a regular hexagon, the diagonals can be expressed in terms of the sides. Specifically:
- The length of diagonal \( AD \) is equal to twice the length of the side \( BC \).
- The length of diagonal \( EB \) is equal to twice the length of the side \( FA \).
- The length of diagonal \( FC \) is equal to twice the length of the side \( AB \).

4. **Expressing the Vectors**:
- We can write:
\[ \vec{AD} = 2 \vec{BC} \]
\[ \vec{EB} = 2 \vec{FA} \]
\[ \vec{FC} = 2 \vec{AB} \]

5. **Adding the Vectors**:
Now we can sum these vectors:
\[ \vec{AD} + \vec{EB} + \vec{FC} = 2 \vec{BC} + 2 \vec{FA} + 2 \vec{AB} \]
Factoring out the 2 gives:
\[ = 2 (\vec{BC} + \vec{FA} + \vec{AB}) \]

6. **Using the Properties of the Hexagon**:
In a regular hexagon, the sum of the vectors from one vertex to the opposite vertex (like \( \vec{BC} + \vec{FA} + \vec{AB} \)) will equal zero because they form a closed shape. Thus:
\[ \vec{BC} + \vec{FA} + \vec{AB} = 0 \]

7. **Final Calculation**:
Therefore:
\[ \vec{AD} + \vec{EB} + \vec{FC} = 2 \times 0 = 0 \]

### Conclusion:
The final result is:
\[ \vec{AD} + \vec{EB} + \vec{FC} = 0 \]